Multiple Choice
Using the Arrhenius Equation, calculate the activation energy (Ea) in kilojoules per mole for a reaction with rate constants of 0.000116 s⁻¹ at 25 °C and 0.305 s⁻¹ at 85 °C. Express your answer to three significant figures.
15. Chemical Kinetics
Arrhenius Equation
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A reaction is followed and found to have a rate constant of 3.36 x 10^4 M^-1 s^-1 at 344 K and a rate constant of 7.69 M^-1 s^-1 at 219 K. Determine the activation energy for this reaction.
A
42.0 kJ/mol
B
58.2 kJ/mol
C
12.5 kJ/mol
D
23.8 kJ/mol
Verified step by step guidance1
Identify that the problem involves determining the activation energy using the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and activation energy (Ea). The equation is: k = A * e^(-Ea/(RT)), where A is the pre-exponential factor, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.
Use the Arrhenius equation in its logarithmic form to compare the rate constants at two different temperatures: ln(k2/k1) = (Ea/R) * (1/T1 - 1/T2). This form allows us to solve for the activation energy (Ea) using the given rate constants and temperatures.
Substitute the given values into the equation: k1 = 3.36 x 10^4 M^-1 s^-1 at T1 = 344 K, and k2 = 7.69 M^-1 s^-1 at T2 = 219 K. The gas constant R is 8.314 J/mol·K.
Calculate the natural logarithm of the ratio of the rate constants: ln(k2/k1). This involves taking the natural logarithm of (7.69 / 3.36 x 10^4).
Solve for the activation energy (Ea) by rearranging the equation to Ea = R * ln(k2/k1) / (1/T1 - 1/T2). Substitute the calculated value of ln(k2/k1) and the temperatures (in Kelvin) into this equation to find Ea in J/mol, then convert to kJ/mol by dividing by 1000.
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